A $C^2$ piecewise Hermite interpolant and a cubic spline are one and the same!
Remember what's done to derive the tridiagonal system: we require that at a joining point, the second left derivative and the second right derivative should be equal.
To that end, consider the usual form of a cubic Hermite interpolant over the interval $(x_i,x_{i+1})$:
$$y_i+y_i^{\prime}\left(x-x_i\right)+c_i\left(x-x_i\right)^2+d_i\left(x-x_i\right)^3$$
where
$$\begin{align*}c_i&=\frac{3s_i-2y_i^\prime-y_{i+1}^\prime}{x_{i+1}-x_i}\\
d_i&=\frac{y_i^\prime+y_{i+1}^\prime-2s_i}{\left(x_{i+1}-x_i\right)^2}\\
s_i&=\frac{y_{i+1}-y_i}{x_{i+1}-x_i}\end{align*}$$
and $\{y_i^\prime,y_{i+1}^\prime\}$ are the slopes (derivative values) of your interpolant at the corresponding points $(x_i,y_i)$, $(x_{i+1},y_{i+1})$.
Take the second derivative of the interpolant over $(x_{i-1},x_i)$ evaluated at $x=x_i$ and the second derivative of the interpolant over $(x_i,x_{i+1})$ evaluated at $x=x_i$ and equate them to yield (letting $h_i=x_{i+1}-x_i$):
$$c_{i-1}-c_i+3d_{i-1}h_{i-1}=0$$
Replacing $c$ and $d$ with their explicit expressions and rearranging yields:
$$h_i y_{i-1}^{\prime}+2(h_{i-1}+h_i)y_i^{\prime}+h_{i-1} y_{i+1}^{\prime}=3(h_i s_{i-1}+h_{i-1} s_i)$$
which can be shown to be equivalent to one of the equations of your tridiagonal system when $h$ and $s$ are replaced with expressions in terms of $x$ and $y$.
Of course, one could instead consider the cubic interpolant in the following form:
$$y_i+\beta_i\left(x-x_i\right)+\frac{y_i^{\prime\prime}}{2}\left(x-x_i\right)^2+\delta_i\left(x-x_i\right)^3$$
where
$$\begin{align*}\beta_i&=s_i-\frac{h_i(2y_i^{\prime\prime}+y_{i+1}^{\prime\prime})}{6}\\\delta_i&=\frac{y_{i+1}^{\prime\prime}-y_i^{\prime\prime}}{6h_i}\end{align*}$$
Doing a similar operation as was done for the Hermite interpolant to this form (except here, one equates first derivatives instead of second derivatives) yields
$$h_{i-1} y_{i-1}^{\prime\prime}+2(h_{i-1}+h_i)y_i^{\prime\prime}+h_i y_{i+1}^{\prime\prime}=6(s_i-s_{i-1})$$
which may be the form you're accustomed to.
To complete this answer, let's consider the boundary condition of the "natural" spline, $y_1^{\prime\prime}=0$ (and similarly for the other end): for the formulation where you solve the tridiagonal for the second derivatives, the replacement is straightforward.
For the Hermite case, one needs a bit of work to impose this condition for the second derivative. Taking the second derivative of the interpolant at $(x_1,x_2)$ evaluated at $x_1$ and equating that to 0 yields the condition $c_1=0$; this expands to
$$\frac{3s_1-2y_1^\prime-y_2^\prime}{x_2-x_1}=0$$
which simplifies to
$$2y_1^\prime+y_2^\prime=3s_1$$
which is the first equation in the tridiagonal system you gave. (The derivation for the other end is similar.)
Roughly speaking ...
A spline is a curve that's formed from a collection of simple segments strung end-to-end so that their junctions are fairly smooth. There are exotic splines that use trigonometric and hyperbolic functions, but most splines consist of polynomial segments, so those are the only ones considered in the discussion below.
If there is only one (polynomial) segment, the spline is often called a Bézier curve.
If each segment is expressed in Bézier form (using Bernstein basis functions), then you might say that the spline is a "Bézier spline", though this term is not standard, AFAIK.
If each polynomial segment has degree 3, the spline is called a cubic spline.
If each segment is described by its ending positions and derivatives, it is said to be in "Hermite" form.
The b-spline approach gives a way of ensuring continuity between segments. In fact, you can show that every spline can be represented in b-spline form. So, in that sense, every spline is a b-spline.
For more than you would ever want to know about the subject, you can search for "spline" or "b-spline" in this bibliography.
Best Answer
You still can do what you want to do with unequal spacing of X values. For your particular case where you have 5 points, this means you have 4 cubic splines you would like to solve for. Therefore, you have 16 unknowns (each cubic spline has 4 coefficients). You have 5 points and 3 continuity constraints ($C^0$, $C^1$ and $C^2$) at each of the 3 interior data points, which amount to 14 equations. The two additional constraints come from setting the second derivatives to zero at the start and end point. So, at the end you have 16 unknowns with 16 equations and you can get your linear equations set up and solved.
In a nutshell, in the process of setting up these linear equations, whether the X values are spaced evenly or unevenly only affects the 16x16 matrix value, but it will not interfere with how the matrix is set up.